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Check column 600x400 with 8-28 mm bars

Short answer

The 400×600 mm column with 8-28 mm bars is not safe. Under Pu = 2000 kN, Mux = 350 kN·m, Muy = 50 kN·m, the exponent alpha_n = 1.58 and the biaxial interaction sum (350/356.1)^1.58 + (50/213.9)^1.58 = 1.074 exceeds 1.0.

A reinforced concrete column of size 600 mm x 400 mm with AstA_{st} = 8-28 mm diameter is subjected to MuxM_{ux} = 350 kN-m, MuyM_{uy} = 50 kN-m and PuP_u = 2000 kN. Check whether the column is safe for above combination of load and bending moments. Take M20 and Fe415 grades of concrete and steel respectively and the effective cover to reinforcing bar 60 mm.

Data

  • Column 400 × 600 mm, 8-28 mm bars (Asc=4926A_{sc} = 4926 mm², p=2.05%p = 2.05\%), arranged 3 bars on each 400 mm face and 1 bar at mid-height of each side (3-2-3 arrangement). Effective cover d′=60d' = 60 mm, so d′/D=0.10d'/D = 0.10 (major axis) and d′/b=0.15d'/b = 0.15 (minor axis).
  • Pu=2000P_u = 2000 kN, Mux=350M_{ux} = 350 kN·m (about the major axis), Muy=50M_{uy} = 50 kN·m; M20 concrete, Fe415 steel. The column is taken as short (no length is given), and the moments are larger than PueminP_u e_{min} (about 40 kN·m), so they are used as given.

Uniaxial capacities at PuP_u = 2000 kN

PufckbD=200000020×400×600=0.417\dfrac{P_u}{f_{ck} bD} = \dfrac{2000000}{20 \times 400 \times 600} = 0.417 and pfck=0.103\dfrac{p}{f_{ck}} = 0.103. From the SP-16 interaction chart for Fe415 (or equivalently by strain compatibility, computed here):

  • About the major axis: Mux1=356.1M_{ux1} = 356.1 kN·m
  • About the minor axis: Muy1=213.9M_{uy1} = 213.9 kN·m

Biaxial check (IS 456 cl. 39.6)

Puz=0.45fckAc+0.75fyAsc=0.45×20×235074+0.75×415×4926=3649P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 235074 + 0.75 \times 415 \times 4926 = 3649 kN PuPuz=0.548\dfrac{P_u}{P_{uz}} = 0.548, so αn=1.58\alpha_n = 1.58 (linear interpolation between 1.0 at 0.2 and 2.0 at 0.8).

(MuxMux1)αn+(MuyMuy1)αn=(350356.1)1.58+(50213.9)1.58=1.074\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{350}{356.1}\right)^{1.58} + \left(\frac{50}{213.9}\right)^{1.58} = 1.074

Since 1.074 >1.0> 1.0, the column is not safe for the given combination of load and moments; the section or the steel must be increased.

✓ Answer checked by Claude Opus 5.5 (high effort)

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